A. DIY Wooden Ladder

Let's denote a kk-step ladder as the following structure: exactly k+2k+2 wooden planks, of which

  • two planks of length at least k+1k+1 — the base of the ladder;
  • kk planks of length at least 11 — the steps of the ladder;

Note that neither the base planks, nor the steps planks are required to be equal.

For example, ladders 11 and 33 are correct 22-step ladders and ladder 22 is a correct 11-step ladder. On the first picture the lengths of planks are [3,3][3,3] for the base and [1][1] for the step. On the second picture lengths are [3,3][3,3] for the base and [2][2] for the step. On the third picture lengths are [3,4][3,4] for the base and [2,3][2,3] for the steps.

You have nn planks. The length of the ii-th planks is aiai. You don't have a saw, so you can't cut the planks you have. Though you have a hammer and nails, so you can assemble the improvised "ladder" from the planks.

The question is: what is the maximum number kk such that you can choose some subset of the given planks and assemble a kk-step ladder using them?

Input

The first line contains a single integer TT (1≤T≤1001≤T≤100) — the number of queries. The queries are independent.

Each query consists of two lines. The first line contains a single integer nn (2≤n≤1052≤n≤105) — the number of planks you have.

The second line contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤1051≤ai≤105) — the lengths of the corresponding planks.

It's guaranteed that the total number of planks from all queries doesn't exceed 105105.

Output

Print TT integers — one per query. The ii-th integer is the maximum number kk, such that you can choose some subset of the planks given in the ii-th query and assemble a kk-step ladder using them.

Print 00 if you can't make even 11-step ladder from the given set of planks.

Example

input

4
4
1 3 1 3
3
3 3 2
5
2 3 3 4 2
3
1 1 2

output

2
1
2
0

代码：

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define inf 0x3f3f3f3f
#define mem(a,b) memset(a,b,sizeof(a))

int a[100005]; 

int main()
{
	int t,n,i;
	cin>>t;
	while(t--)
	{
		cin>>n;
		for(i=0;i<n;i++)
			cin>>a[i];
		sort(a,a+n);
		int t=a[n-2];
		n-=2;
		if(t>n)
			cout<<n<<endl;
		else
			cout<<t-1<<endl;
	} 
	return 0;
}